CAIE M1 2023 June — Question 5 9 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2023
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeEquilibrium of particle under coplanar forces
DifficultyStandard +0.3 This is a standard equilibrium problem requiring resolution of forces in two perpendicular directions. Part (a) involves setting up two simultaneous equations (ΣFx=0, ΣFy=0) and solving for F and θ—routine for M1 students. Part (b) is straightforward vector addition with given values. While it requires careful trigonometry and multi-step working, it follows standard textbook methods with no novel insight needed, making it slightly easier than average.
Spec3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03p Resultant forces: using vectors
\includegraphics{figure_5} Four coplanar forces act at a point. The magnitudes of the forces are \(F\) N, \(10\) N, \(50\) N and \(40\) N. The directions of the forces are as shown in the diagram.
  1. Given that the forces are in equilibrium, find the value of \(F\) and the value of \(\theta\). [6]
  2. Given instead that \(F = 10\sqrt{2}\) and \(\theta = 45\), find the direction and the exact magnitude the resultant force. [3]
Question 5:
AnswerMarks
5Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
AnswerMarks Guidance
5(a)Resolving either direction. M1
be an equation with either = 0 or with an attempt to
balance forces.
AnswerMarks Guidance
Vertical: Fsin40sin60500A1 Fsin5020 315.358...
 
AnswerMarks Guidance
Horizontal: Fcos1040cos600A1 Fcos10
tan1(52 3)M1 Attempt to solve for ; one missing term in total
tan11.535898....
AnswerMarks Guidance
F  15.358...2 102M1 Attempt to solve for F: one missing term in total.
θ = 56.9, F = 18.3A1 Both correct (18.327530…, 56.932462…).
6
AnswerMarks Guidance
5(b)(Y )(10 2sin4540sin6050)[(20 340)] B1
expression.
AnswerMarks Guidance
(X )(10 2cos451040cos60)[0]B1 Allow non-exact values for 2 etc. in correct
expression. Could be implied by correct answer.
AnswerMarks Guidance
Resultant force is 4020 3 (N) in the same direction as the 50 (N) force.B1 Allow vertically downwards, south, 180, negative y-
direction.
Resultant force must be exact and positive (so
20 340is B0).
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | Resolving either direction. | M1 | 3 terms; allow sign errors and allow sin/cos mix. Must
be an equation with either = 0 or with an attempt to
balance forces.
Vertical: Fsin40sin60500 | A1 | Fsin5020 315.358...
 
Horizontal: Fcos1040cos600 | A1 | Fcos10
tan1(52 3) | M1 | Attempt to solve for ; one missing term in total
tan11.535898....
F  15.358...2 102 | M1 | Attempt to solve for F: one missing term in total.
θ = 56.9, F = 18.3 | A1 | Both correct (18.327530…, 56.932462…).
6
--- 5(b) ---
5(b) | (Y )(10 2sin4540sin6050)[(20 340)] | B1 | Allow non-exact values for 2 etc. in correct
expression.
(X )(10 2cos451040cos60)[0] | B1 | Allow non-exact values for 2 etc. in correct
expression. Could be implied by correct answer.
Resultant force is 4020 3 (N) in the same direction as the 50 (N) force. | B1 | Allow vertically downwards, south, 180, negative y-
direction.
Resultant force must be exact and positive (so
20 340is B0).
3
Question | Answer | Marks | Guidance
\includegraphics{figure_5}

Four coplanar forces act at a point. The magnitudes of the forces are $F$ N, $10$ N, $50$ N and $40$ N. The directions of the forces are as shown in the diagram.

\begin{enumerate}[label=(\alph*)]
\item Given that the forces are in equilibrium, find the value of $F$ and the value of $\theta$. [6]
\item Given instead that $F = 10\sqrt{2}$ and $\theta = 45$, find the direction and the exact magnitude the resultant force. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2023 Q5 [9]}}
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